I sort of feel like I deal with trolls and I hope that is not the case.

What are you afraid of? If your claims are sound, you would instantly know the answer for the questions we've posed. Even for the one Sredni found amusing. If you think that that question is out of proportion, think again. That could very well be traces on a PCB or wires on any real installation. If your Kirchhoff only works for perfectly rectangular or round loops, with known resistors within tolerances, it is a useless theory.

And here is one more challenge for your Kirchhoff. You know the drill. We need to know the correct way to measure the correct voltage on this loop. The B field is now concentrated on half of the area, but all other conditions hold. Good luck.

I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .I can tell you that in my case it would be in the mV/m range. What value do you get in your case?

OK I think I understand what part you do not understand so I will try to find a bit of a different example to explain the problem.

Imagine you have a ring (closed loop) made of superconducting material (set flat on a table) and somewhere above it you have a magnet.As long as magnet and ring are stationary there will be no EMF so of course also no current in that ring.Now I drop that magnet that will start to accelerate thus there is a change in magnetic field and there will be a current induced inside the ring that will create a field that opposes the magnetic field from the magnet.This opposing magnetic field will slow down the magnet and in this particular case it will be slowed down to zero thus you will end up with a levitating magnet.Now what happened is that energy from the falling magnet induced a current in the coil that created an opposing magnetic field and since coil is made of superconducting material there is no IR loss as R=0 so that energy remains conserved (you can even slowly remove the magnet and that current will remain in that ring).

If we repeat the experiment but instead of the superconducting ring we now use a copper conductor but say is fairly think very low resistance the magnet will still slow down but since there is a bit of IR loss magnet will not levitate and fall all the way down until will hit the table or floor.

Now third experiment you can have a copper cor super conductor ring but it will be an open loop so a very small small cut and you can install a voltmeter there but should be with almost infinite impedance.In this case magnet will not slow down at all (we ignore the air resistance) but voltmeter will read a voltage that is basically the EMF as IR is zero I=0

EMF = Velocity x B-flux x length so if you want to increase EMF you will need to increase the speed in this case you need to apply some additional force to the magnet or drop it from a higher place or increase the length of the loop but then you also need to increase the size of the magnet so B-flux remains the same.

We never mentioned in earlier examples the B-flux or the length of our loop as we had enough other information like we established that current trough the loop was 1A and the loop resistance was 1Ohm this was to have small round numbers.

All this discussion is off topic and not sure it helped but the the Lewin claim was that at the same moment in time you can have two completely different voltages on the same exact two points. That is just bad measurement methods as he did not considered the voltmeter leads as part of the circuit.

What are you afraid of? If your claims are sound, you would instantly know the answer for the questions we've posed. Even for the one Sredni found amusing. If you think that that question is out of proportion, think again. That could very well be traces on a PCB or wires on any real installation. If your Kirchhoff only works for perfectly rectangular or round loops, with known resistors within tolerances, it is a useless theory.

I'm afraid of you being a troll and wasting my time.Read the replay above as it may be relevant if you are not a troll.Shape of the loop will make no difference as long as B-flux is uniform but if that is not the case then you will not be able to calculate that with just pen and paper (you may be able to approximate something) but you will need a computer simulation tool to solve that and of course all details to scale.And even if flux is uniform you will need to know the total length of that loop and the length between the two points you want to make the measurement then calculation is the same as for the simple ring model as shape alone makes no difference.

I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .I can tell you that in my case it would be in the mV/m range. What value do you get in your case?

OK I think I understand what part you do not understand so I will try to find a bit of a different example

Can you or can you not compute this electric field?Our circuit is stationary, nothing is moving.What is the electric field inside your conductor?

You are not telling because you do not know how to compute it, or because you cannot justify a 0.25 V (yeah, minus 0.002V) voltage difference at its ends?Because that's the point.

Can you or can you not compute this electric field?Our circuit is stationary, nothing is moving.What is the electric field inside your conductor?

You are not telling because you do not know how to compute it, or because you cannot justify a 0.25 V (yeah, minus 0.002V) voltage difference at its ends?Because that's the point.

This tread is here to discuss about the silly idea that you can read two different voltages measured between the exact same two points at the exact same moment in time.

But good news .I came up with an experiment that you can do to understand EMFIt involves either a transformer or if you prefer a brush-less motor/generator (maybe you have one small servo motor around).

Say it is the servo motor as it is more visual and shows a bit more information.EMF output 100V at 1000RPM and coil resistance 10OhmYou just connect a voltmeter (you can ignore the 1Mohm) and while the generator spins at 1000RPM you will read 100VNow you connect at 100Ohm restive load and again spin at 1000RPM what do you think the voltmeter will read now at the output terminal (in parallel with the 100Ohm load) ?It will of course take more work to spin at 1000RPM when the 100Ohm load is connected but the EMF will be the same 100V

So you have 100V and the close loop will have 100 + 10 = 110Ohm thus current will be 0.909AIR Voltage drop on the internal 10Ohm coil will be 9.09V thus the voltmeter will read 100V - 9.09V = 90.9V

If you do not believe my calculations are correct then you can experiment. Fortunately the motor and transformer are both shielded so it will not mess with your measurement device.

Edit: This was my last attempt as there is not much else I can do.I have to go back at creating equipment's that work based in part on what I tried to explain above. They are all open source as I think knowledge should be free and shared, and people should understand how things work.

The last two pages of arguing in this thread seam to be all because of different definitions of voltage.

The correct formal definition used by Dr Lewin: Voltage is the integral of all forces on charges along a given path

The common definition used in circuit analysis: Voltage is the difference in charge density between two points

The formal definition is what results in the two voltages because it handles magnetic EMF differently. The EMF is path dependent and this makes the voltage path depend an as well. Because some part of the path is missing this gives multiple results until you close the path using a voltmeter. In this definition a superconductor can't have any voltage across it.

The other definition just focuses on how many electrons there are in that spot. It essentially ignores any magnetic EMF and instead observes the charge separation effect that the magnetic fields cause. Since there can only be a single number for how many electrons are there means this never gives multiple voltages as a result. In this definition superconductors can have a voltage across them if they are a open loop, the magnetic forces bunch up the electrons to one end of the wire and you can measure this voltage with a voltmeter. This is because voltmeters read this definition of voltage. I call this "aparrant voltage"

So I suggest that you make it clear in further discussion what kind of voltage you are talking about.

Far to much theoretical waffle and going around in circles or in some cases non circular circuits. Given this is a forum for EE's not theoretical Physicists with very average probing technique, get on design the experiment, test it and prove or disprove it and then get it repeated by others.

I'm still working on an openEMS simulation. Way too many hours into this now... Need a faster computer. Not done yet, but I have run across something that I found interesting. I'm trying to follow Romer's paper now (wish I had looked at that as soon as I ran across this thread). I have to make some compromise because I don't have a super-computer. Based on the paper, the important thing is that the current is changing linearly while the measurements are made. I have the model pretty close, but less turns on the solenoid and somewhat higher frequency triangle wave. The feed current very high just to scale things up. Here is the interesting part. I have been seeing a high frequency component on one side relative to the other. I realized that this is being caused by the H-field given off by the solenoid feed being closer to one side of the loop than the other. Also looks like the high frequency could be partially some interaction between the solenoid turns / feedline. Note I am using 1k and 2k resistor values as romer does, but look at the voltmeter (integrated current across 20meg resistors) at the beginning of the sim. The higher voltage is about 10x greater (look familiar?). It also rises before the other side because of the asymmetry caused by the feedline being closer to that side. Current feed is nice and linear during these plots. The geometry is symmetrical except for the feedline which brings up a question about all of the physical experiments that have been done. How would you go about getting rid of the effects of the feedline in a real-world experiment? I'm no mathematician (obviously), but doesn't this make it kind of hard to get a symmetrical field? Romer seems to hint at this: " Note that to the extent that the magnetic field exterior to the solenoid can be neglected....". Should we add bad sourcing to the list along side of bad probing?

You agreed with ogden, who expressed electric field in volts. You should have read better, before saying that his reply was more to the point.

Well, ok. I admit that ogden (*me) did not answer your trolling question "what is the field inside the copper conductor" correctly. Expressing Volts, I missed to mention "integral of E.dl". Hopefully it is resolved now and we can return to the roots of our conversation.

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What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?

BTW wire fragment resistance is 0.001 Ohms so it is (0.25-0.001) Volts, not (0.25-0.002).

Inside our conductor there are two E fields: E.induced and E.coloumb. Total electric field E = E.coloumb + E.induced. Coulomb electric field in the wire is opposite the direction of the induced electric field - that's the justification of voltage difference. Potential difference (integral of E.dl) at the ends of that copper conductor you calculate using same formula as for 0.25V chemical battery that has 0.001 Ohm internal resistance and 1A current load. Answer is mentioned already here in this thread.

What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?

BTW wire fragment resistance is 0.001 Ohms so it is (0.25-0.001) Volts, not (0.25-0.002).

Ok.

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Inside our conductor there are two E fields: E.induced and E.coloumb. Total electric field E = E.coloumb + E.induced. Coulomb electric field in the wire is opposite the direction of the induced electric field

Ok, this is real progress.

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- that's the justification of voltage difference.

Please clarify.The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which? The copper sees the net, resulting, field. (and THIS is the point)What is this sum?

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Potential difference (integral of E.dl) at the ends of that copper conductor you calculate using same formula as for 0.25V chemical battery that has 0.001 Ohm internal resistance and 1A current load. Answer is mentioned already here in this thread.

Please, indulge me. Give me the number in V/m (volts per meter).

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Assume standard conductivity for copper, say 5.8 10^7 mhos per meter

Do not introduce new unnecessary conditions.

I am looking forward to seeing how you compute the field and how well it fits with the 0.25-0.001 volts difference at the extremes.

The last two pages of arguing in this thread seam to be all because of different definitions of voltage.

The correct formal definition used by Dr Lewin: Voltage is the integral of all forces on charges along a given path

The common definition used in circuit analysis: Voltage is the difference in charge density between two points

[...]So I suggest that you make it clear in further discussion what kind of voltage you are talking about.

Just out of curiosity...Have you ever heard the term "dimensional analysis"?

(Also, I guess you won't tell us what the field inside the conductor is... Kirchhoffians are allergic to electric fields, it appears)

I am not sure I can sort this mess.

Yes i have its getting units in order in equations.

What so problematic about the field? You get a electric field caused by charge separated electrons that is precisely proportional to the amount of charge separation. There is also an apparent electric field caused by the magnetic field that is exactly the same size and opposite in direction than the electric field from before (Given this conductor is an open loop or a superconductor)

I'd like to kindly ask the Kirchhoff experts what tools do Kirchhoff rules give me to calculate the "right" voltage of the loop below. Except for the irregular perimeter, all conditions are the same as for my rectangular loop above.

I need to know the exact points where to connect the voltmeter and its precise location. Any reply will be appreciated.

Oh and also i solved your curvy wire example.

Here is where you have to put the voltmeter for it to read 0V

Solved using Solidworks Fusion 360 due to it having a convenient area measurement tool.

The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which?

What are you smoking?

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The copper sees the net, resulting, field. (and THIS is the point)What is this sum?

Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?

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Potential difference (integral of E.dl) at the ends of that copper conductor you calculate using same formula as for 0.25V chemical battery that has 0.001 Ohm internal resistance and 1A current load. Answer is mentioned already here in this thread.

[p.s.] Trolls... One is shifting goalposts all over the place, I wonder when he will touch chemical composition of the copper wire? Another is stuck into self-inflicted mental loop of proving that Kirchoff's rules cannot be used in place of Maxwells equations. Fun!

Solved using Solidworks Fusion 360 due to it having a convenient area measurement tool.

Wonderful. Much appreciated. Now we need to measure the voltage indicated by the calculations so as to confirm that they are right. But, alas, in the real circuit there is a physical obstruction, that in no way affects the magnetic field. This obstruction goes all the way with the field while it is perpendicular to the surface.

How can we measure measure that voltage? Thanks in advance for your kind reply.

Far to much theoretical waffle and going around in circles or in some cases non circular circuits. Given this is a forum for EE's not theoretical Physicists with very average probing technique, get on design the experiment, test it and prove or disprove it and then get it repeated by others.

By the choice of your words you sense that there is probably something wrong with your "probing technique". It's not sponsored by any electronics engineering fundamentals which pretty much describes tried and true experimental phenomena along the past two centuries up to this day. You only rely on a couple of 10 min or so videos on the internet without even questioning their content. Any serious trade like ours upon which the lives of people depend deserves a little more rigor.

Solved using Solidworks Fusion 360 due to it having a convenient area measurement tool.

Wonderful. Much appreciated. Now we need to measure the voltage indicated by the calculations so as to confirm that they are right. But, alas, in the real circuit there is a physical obstruction, that in no way affects the magnetic field. This obstruction goes all the way with the field while it is perpendicular to the surface.

How can we measure measure that voltage? Thanks in advance for your kind reply.

That's just an engineering problem at this point.

1) Make a length of wire that would fit across those two points as if the pesky barrier was not there2) Make a rigid wire structure that goes around the barrier as needed and connects to a voltmeter on the other end3) Make another copy of the structure from 2, but short it using the piece of wire from 14) Place the structure from 2 onto the circuit to tap the voltage and place the structure from 3 anywhere near by5) Subtract the readings of the voltmeters.

The compensation structure from 3 can be used multiple times to ensure the field is indeed uniform all around so that we know the placement of the structure has shown valid readings.

Alternatives are to just calculate the voltage of the compensation structure if you already know the exact properties of the field, or in that case if you know the properties of the field and the path of the wire you want to measure you can just apply Faradays law to the whole thing.

Remember im not trying to disprove anything about Faraday or Maxwell. Just saying that Kirchhoffs laws are convenient to use in some cases (And they do work when used correctly).

Wonderful. Much appreciated. Now we need to measure the voltage indicated by the calculations so as to confirm that they are right. But, alas, in the real circuit there is a physical obstruction, that in no way affects the magnetic field. This obstruction goes all the way with the field while it is perpendicular to the surface.

How can we measure measure that voltage? Thanks in advance for your kind reply.

Just to be sure rules are known - both resistors are equal and positioned symmetrically against midpoint of the the loop, right?

Sure - due to obstruction and EMF we can't measure potential difference between given points directly. First we measure EMF induced in the voltmeter test leads - by shorting them on far side and routing them around the obstruction, making sure our test lead loop is symmetrical and centered against outer loop. Then we can leave one lead where it is (connected to far side midpoint of the outer loop) and bring another to near midpoint, measure voltage across the outer loop noting that it is impaired by EMF in one of test leads. Then just either add or subtract 1/2 of test leads EMF voltage from measurement - depending on which test lead receives EMF and direction of magnetic field. What's the point of all this?

[p.s.] Let's name shape of this loop or maybe even whole experiment as "trail of the troll"...

The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which?

What are you smoking?

ogden, I promised not to interact with you, but you are making this promise very hard to keep. If you interfere with my exchanges with other posters I have to reply to you as well.

Quote from: ogden

Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?

Because you guys keep telling me the voltage. I want to know the electric field.

So, let me ask you again, because this is important:What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?

Assume standard conductivity for copper, say 5.8 10^7 mhos per meter and a copper section of 1 mm in diameter (or any real world value you can attribute to a circuit similar to those shown by Lewin, Mehdi or Mabilde - it's about 10 cm diameter loop, suppose half of it is allocated by the big resistors, but it's not important).

I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .I can tell you that in my case it would be in the mV/m range. What value do you get in your case?

Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?

Because you guys keep telling me the voltage. I want to know the electric field.

I said (E = E.coloumb + E.induced). Are you satisfied now?

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Quote from: Sredni

Assume standard conductivity for copper, say 5.8 10^7 mhos per meter and a copper section of 1 mm in diameter (or any real world value you can attribute to a circuit similar to those shown by Lewin, Mehdi or Mabilde - it's about 10 cm diameter loop, suppose half of it is allocated by the big resistors, but it's not important).

I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .I can tell you that in my case it would be in the mV/m range. What value do you get in your case?

And still, no answer.

You can either provide solution yourself and tell what you want to say with it or stick that tangential component where it hurts. I do not see the point of solving your tasks. "Trail of the troll" was at least funny.

Because you guys keep telling me the voltage. I want to know the electric field.

I said (E = E.coloumb + E.induced). Are you satisfied now?

No, I want to know the value in V/m (or in J/C if you prefer).

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You can either provide solution yourself and tell what you want to say with it or stick that tangential component where it hurts. I do not see the point of solving your tasks. "Trail of the troll" was at least funny.

Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine. Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?